Mathematics · Quantitative Aptitude

Sequences and Series

226 Questions

Sequences and series involve ordered lists of numbers and the sum of their terms. The questions primarily test knowledge of arithmetic progressions, geometric progressions, and infinite series. It is an essential part of quantitative aptitude that requires strong pattern recognition skills.

Arithmetic progressionGeometric progressionInfinite seriesSum of termsNumber sequences

Sequences and Series Questions

Multiple choice

Find the 19th term of the given squence. 10, 14, 19, 25...

  1. 235

  2. 236

  3. 248

  4. 259
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

nth term = a + (n – 1)d + ${\frac 12}$ (n – 1) (n – 2) c a = 10, d = 4, c = 1, n = 19               = 10 + (19 – 1) 4 + ${\frac 12}(19-1)(19-2)$1              = 10 + 18 $\times$ 4 + 9 $\times$ 17              = 235

Multiple choice

Which term of the AP: 3, 8, 13, 18, _______ is 78?

  1. 15th term

  2. 16th term

  3. 17th term

  4. 14th term
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

 a = 3 and d = 5 (18 - 13, 13 - 8, 8 - 3 = 5) 78 = an = a + (n - 1)d 78 = 3 + (n - 1)5 78 = 3 + 5n - 5 $\Rightarrow$ 78 = 5n - 2 $\Rightarrow$ 78 + 2 = 5n 80 = 5n $\Rightarrow$ n = 16. So, 78 is the 16th term of the AP.

Multiple choice

The sum of n terms of the series 4 + 44 + 444 +.... is

  1. (4/81) [10n+1 - 9n - 1]

  2. (4/81) [10n-1 - 9n - 1]

  3. (4/81) [10n+1 - 9n - 10]

  4. (4/81) [10n - 9n - 10]
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The series 4 + 44 + 444 + ... can be rewritten as (4/9) * (9 + 99 + 999 + ...), which becomes (4/9) * [(10 - 1) + (10^2 - 1) + ... + (10^n - 1)]. Summing the geometric progression and the constant term yields the standard closed-form expression (4/81) * [10^(n+1) - 9n - 10].

Multiple choice statistics normal distribution distribution of measurement probability distributions introduction to normal distribution

$\sum _{r=1}^{11} r.5^{r} =\dfrac{(43\times 5^{a}+5)}{b}$, then $(a+b)$ is

  1. $18$
  2. $28$
  3. $15$
  4. $38$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Evaluating the arithmetico-geometric series sum using standard methods yields the specified numerator and denominator form. Matching coefficients and powers gives the values leading to a sum of 28 for a plus b.

Multiple choice median percentiles and quartiles range and mean deviation mode

If the 2nd term of a GM series is 25, the first and 3rd terms will be _____.

  1. 5,125

  2. 105,5

  3. 5,105

  4. 25,125
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Geometric mean between two numbers is the mean proportion of the series. 

Therefore, let the two numbers be a and b 
a/25 = 25/b 

=>ab= 25 x 25

Also , a/b = 1/25

 b = 25a

so, a (25a) = 25 x 25

=>a2 = 25

=> a = 5

Now, b = 25(5)

        b = 125  

Therefore, the first and the 3rd terms are 5 and 125 respectively. 

Multiple choice taylor's and maclaurin's series applications of differential calculus maths

If the sum of the series $\dfrac{3}{1!}+\dfrac{5}{2!}+\dfrac{7}{3!}+\dfrac{9}{4!}+...\infty=Ae+B$
Find the value of $A+B$

  1. $1$
  2. $7$
  3. $0$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Its general term is $\dfrac{2n+1}{n!}$

So we have to calculate $\sum^{n=\infty} _{n=0}\dfrac{2n+1}{n!}=2\sum^{k=\infty} _{k=0}\dfrac{1}{k!}+\sum^{n=\infty} _{n=0}\dfrac{1}{n!}-2=3e-2$ (using  taylor's expansion for $e^x$)

So A+B=1

Multiple choice statistics measures of central tendency geometric and harmonic mean geometric mean mean

Find the sum of 5 geometric means between $\displaystyle\frac{1}{3}$ and 243, by taking common ratio positive.

  1. 121

  2. 126

  3. 81

  4. 111
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given that, there are $5$geometric means between the two numbers $\dfrac{1}{3}$and $243$ , we have to find $7=\left( 5+2 \right)$

terms in G.P. of which $\dfrac{1}{3}$ is the first, and$243$ the seventh. Let r be the common ratio;

then $243$  = the seventh term =$\left( \dfrac{1}{3} \right){{r}^{\left( 7-1 \right)}}=\dfrac{1}{3}.{{r}^{6}}$.

 

Therefore,${{r}^{6}}=3.x.243={{3.3.3}^{4}}={{3}^{6}}$;

whence $r=6$

and the series is$\dfrac{1}{3},1,3,9,27,81,243$

(using the standard form a, ar, ar², ar³ …… of a G.P. ).

 

Now, the geometric mean between two given quantities$a,b=\sqrt{ab}$

 

Therefore, the required geometric means are,

$ \sqrt{\dfrac{1}{3}.x.3},\sqrt{1.x.9},\sqrt{3.x.27},\sqrt{9.x.82},\sqrt{27.x.243} $$

$ =1,3,9,27,81 $$

 

Therefore, the sum of the $5$  geometric means is

\$1+3+9+27+81=121$

 

Hence, this is the answer.